A) \[{{\left( \frac{AB}{AC} \right)}^{2}}\]
B) \[\frac{AB}{AC}\]
C) \[{{\left( \frac{AB}{AD} \right)}^{2}}\]
D) \[\frac{AB}{AD}\]
Correct Answer: A
Solution :
[a] In a right triangle ABC, right-angled at A and \[AD\bot BC\]. |
In \[\Delta CAB\] and \[\Delta CDA,\] |
\[\angle CAB=\angle CDA\] (Each\[90{}^\circ \]) |
\[\angle ACB=\angle DCA\] (Common) |
\[\therefore \,\,\Delta CAB\tilde{\ }\Delta CDA\](By AA similarity criterion) |
\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\frac{BC}{AC}=\frac{AC}{DC}\] |
\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,A{{C}^{2}}=BC\cdot DC\] ..(1) |
Similarly, \[\Delta CAB\tilde{\ }\Delta ADB\] |
(By AA similarity criterion) |
\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{AB}{BD}=\frac{BC}{AB}\] |
\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,A{{B}^{2}}=BC\cdot BD\] .(2) |
Dividing (2) by (1), we set |
\[\frac{A{{B}^{2}}}{A{{C}^{2}}}=\frac{BC\cdot BD}{BC\cdot DC}=\frac{BD}{DC}\] |
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